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Under what circumstance does the integral x0xbdxdiverge? Use this to argue that a physically acceptable wave function must fall to 0 faster than|x|1/2 does as xgets large.

Short Answer

Expert verified

The function will keep on diverging unless b is less than hence we can say the integral will diverge at the given infinite limit (upper limit). The value of b equal to 1 cannot be possible because it would make the entire function infinite. Now, for the total integral to not diverge, it is necessary for the2 to fall off more rapidly or at least as rapid as |x|1 and to fall off more rapidly or at least as rapid as |x|1/2

Step by step solution

01

Given expression 

The Integral is given as,

x0xbdx.

02

Concept of Integration 

Integration concept used

abf(x)dx=F(b)F(a)
03

Integration concept used

In order to calculate the conditions under which the given integral may diverge, start by writing the integral as follows:

x0xbdx

Here, dx is called the differential of the variablex.

The upper limit of the Integral is 8.

04

Integrating the given function

Integrating the above-given function under the given limits we will get

x0xbdx=[xb+1b+1]x0

The function will keep on diverging unless b is less than -1 hence we can say the integral will diverge at the given infinite limit (upper limit). The value of b equal to -1 cannot be possible because it would make the entire function infinite.

Now, for the total integral to not diverge, it is necessary for the 2 to fall off more rapidly or at least as rapid as |x|1 and to fall off more rapidly or at least as rapid as|x|1/2 .

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Most popular questions from this chapter

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It is possible to take the finite well wave functions further than (21) without approximation, eliminating all but one normalization constant C . First, use the continuity/smoothness conditions to eliminate A, B , andG in favor of Cin (21). Then make the change of variables z=x-L/2 and use the trigonometric relations

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functions in region I, -L/2<z<L/2. The change of variables shifts the problem so that it is symmetric about z=0, which requires that the probability density be symmetric and thus that (z)be either an odd or even function of z. By comparing the region II and region III functions, argue that this in turn demands that (/k)sinkL+coskL must be either +1 (even) or -1 (odd). Next, show that these conditions can be expressed, respectively, as k=tankL2 and k=-cotkL2. Finally, plug these separately back into the region I solutions and show that

(z)=C{e(z+L/2)鈥夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌z<L/2coskzcoskL2鈥夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌-L/2<z<L/2e-(z-L/2)鈥夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌z>L/2


or

(z)=C{e(z+L/2)鈥夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌z<L/2-sinkzsinkL2鈥夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌-L/2<z<L/2e-(z-L/2)鈥夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌夆赌z>L/2

Note that Cis now a standard multiplicative normalization constant. Setting the integral of |(z)|2 over all space to 1 would give it in terms of kand , but because we can鈥檛 solve (22) exactly for k(or E), neither can we obtain an exact value for C.

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